# Prove a function is Riemann integrable on an interval - Homework

Let $f_n : [a,b] \rightarrow \Bbb R$ be a sequence of bounded Riemann integrable functions on $[a,b]$ and let $f : [a,b] \rightarrow \Bbb R$ be a bounded function such that $\lim_{n\to\infty} d(f_n,f) = 0$ where $d(f_n,f)=sup[|f_n(x)-f(x)|:a \le x \le b]=0.$

Prove that $f$ is Riemann integrable on $[a,b]$ and that

$\int_{a}^{b}fdx = \lim_{n\to\infty} \int_{a}^{b}f_ndx$.

I need to show this using the Riemann integrability criterion, that for every $\epsilon \gt 0, \exists$ a partition $\mathcal P$, s.t. $U(f,\mathcal P) - L(f,\mathcal P) \lt \epsilon.$

Given $\epsilon$, there is $N$ such that $d(f_n,f)<\frac{(b-a)\epsilon}3$ for all $n>N$. For each such $n$ and any $\mathcal P$, conclude that $U(f,\mathcal P)<U(f_n,\mathcal P)+\frac\epsilon 3$ and $L(f,\mathcal P)>L(f_n,\mathcal P)-\frac\epsilon 3$. Hence for suitable $\mathcal P$ (namely with $U(f_n,\mathcal P)-L(f_n,\mathcal P)<\frac\epsilon 3$) we find that $U(f,\mathcal P)-L(f,\mathcal P)<\epsilon$. This shows integrability. Moreover, $$L(f_n,\mathcal P)-\frac\epsilon 3<L(f,\mathcal P) \le \int_a^bf(x)\,\mathrm dx\le U(f,\mathcal P)<U(f_n,\mathcal P)+\frac\epsilon 3$$ and $$L(f_n,\mathcal P) \le \int_a^bf_n(x)\,\mathrm dx\le U(f_n,\mathcal P)$$ imply that $$\left|\int_a^bf(x)\,\mathrm dx-\int_a^bf_n(x)\,\mathrm dx\right|<U(f_n,\mathcal P)-L(f_n,\mathcal P)+\frac\epsilon 3<\frac23\epsilon$$ and hence $$\int_a^bf(x)\,\mathrm dx=\lim_{n\to\infty}\int_a^bf_n(x)\,\mathrm dx.$$